Question

Find the sum to n terms of the series: $5 + 11 + 19 + 29 + 41 + ...........$

Answer 1

Hint: This is not an AP or GP. It is a special series. Add 0 in the first term then subtract new series with the given series. You will find an AP is produced and move along with it and you will find the required solution.
Complete step by step answer:
Let

{S_n} = 5 + 11 + 19 + 29 + 41 + ...... + {a_{n - 1}} + {a_n}\\\ {S_n} = 0 + 5 + 11 + 19 + 29 + 41 + ....... + {a_{n - 2}} + {a_{n - 1}} + {a_n} \end{array}$$ Now subtract both of them we will get, $$\begin{array}{l} \Rightarrow {S_n} - {S_n} = 5 - 0 + [(11 - 5) + (19 - 11) + (29 - 19) + ....({a_{n - 1}} - {a_{n - 2}}) + ({a_n} - {a_{n - 1}})] - {a_n}\\\ \Rightarrow 0 = 5 + [6 + 8 + 10 + 12 + .....{a_{n - 1}}] - {a_n}\\\ \Rightarrow {a_n} = 5 + [6 + 8 + 10 + 12 + .....{(n - 1)^{th}}term]...............................(i) \end{array}$$ Now it is clearly seen that $$6,8,10,12,.....{(n - 1)^{th}}term$$ We know that the general equation of an AP is given by $${a_n} = a + (n - 1)d$$ Where a is the first term, n is the number of term and d is the common difference Here in this case $$a = 6,d = 2$$ We know that sum of n terms of an AP is given by $$\dfrac{n}{2}\\{ 2a + (n - 1)d\\} $$ Now putting $$n = n - 1,a = 6\& d = 2$$ we get $$\begin{array}{l} = \dfrac{{n - 1}}{2}[2(6) + \\{ (n - 1) - 1\\} 2]\\\ = \dfrac{{n - 1}}{2}[12 + (n - 1 - 1)2]\\\ = \dfrac{{n - 1}}{2}[12 + (n - 2)2]\\\ = \dfrac{{n - 1}}{2}[12 + 2n - 4]\\\ = \dfrac{{n - 1}}{2}[8 + 2n]\\\ = \dfrac{{n - 1}}{2} \times 2[4 + n]\\\ = (n - 1)(n + 4) \end{array}$$ Now we can put this in equation (i) we get $$\begin{array}{l} \Rightarrow {a_n} = 5 + (n - 1)(n + 4)\\\ \Rightarrow {a_n} = 5 + n(n + 4) - (n + 4)\\\ \Rightarrow {a_n} = 5 + {n^2} + 4n - n - 4\\\ \Rightarrow {a_n} = {n^2} + 3n + 1 \end{array}$$ Now, $$\begin{array}{l} {S_n} = \sum\limits_{n = 1}^n {{a_n}} \\\ \Rightarrow {S_n} = \sum\limits_{n = 1}^n {{n^2} + 3n + 1} \\\ \Rightarrow {S_n} = \sum\limits_{n = 1}^n {{n^2}} + \sum\limits_{n = 1}^n {3n} + \sum\limits_{n = 1}^n 1 \\\ \Rightarrow {S_n} = \sum\limits_{n = 1}^n {{n^2}} + 3\sum\limits_{n = 1}^n n + \sum\limits_{n = 1}^n 1 \end{array}$$ We know that sum of all $$n \& {n^2}$$ terms is given by $$\dfrac{{n(n + 1)}}{2}\& \dfrac{{n(n + 1)(2n + 1)}}{6}$$ and also sum of n 1’s will naturally be n only. $$\begin{array}{l} \therefore \dfrac{{n(n + 1)(2n + 1)}}{6} + \left( {\dfrac{{n(n + 1)}}{2}} \right) + n\\\ = \dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{3}{2}n(n + 1) + n\\\ = \dfrac{{n(n + 1)(2n + 1) + 9(n + 1) + 6n}}{6}\\\ = n\left( {\dfrac{{(n + 1)(2n + 1) + 9(n + 1) + 6}}{6}} \right)\\\ = n\left( {\dfrac{{2{n^2} + 2n + n + 1 + 9n + 9 + 6}}{6}} \right)\\\ = n\left( {\dfrac{{2{n^2} + 12n + 16}}{6}} \right)\\\ = n\left( {\dfrac{{2({n^2} + 6n + 8)}}{6}} \right)\\\ = \dfrac{n}{3}({n^2} + 6n + 8)\\\ = \dfrac{n}{3}[n(n + 4) + 2(n + 4)]\\\ = \dfrac{n}{3}[(n + 2)(n + 4)]\\\ = \dfrac{{n(n + 2)(n + 4)}}{3} \end{array}$$ Therefore the required sum is $${\dfrac{{n(n + 2)(n + 4)}}{3}}$$ Note: The key step here was to find the AP from the given series and also remember the summation of $$n,{n^2}\& {n^3}$$ which is $$\dfrac{{n(n + 1)}}{2},\dfrac{{n(n + 1)(2n + 1)}}{6}\& {\left[ {\dfrac{{n(n + 1)}}{2}} \right]^2}$$ respectively. These three are often used while trying to solve problems like this.